Integral of the Von karman equation What is the result of this integral, and how can I proceed:

$$
\int_{-\infty}^{\infty}{c_{1} \over\left(1 + c_{2}\,x^{2}\right)^{5/6}}\,
\cos\left(x\tau\right)\,{\rm d}x\,,\qquad c_{1}, c_{2}\mbox{: positive constants.}
$$

I see in one book that the result contains the
${\tt 2_{\rm nd}\ \mbox{type Bessel function}}$.
 A: Related problems (I). Follow the steps
1) the integrand is even so you can integrate on the interval $(0,\infty)$
2) expand the function $\cos(x)$ in terms of its Taylor series
3) use the beta function to evaluate the integral and then  resum.
Added: Here is a closed form for a more general integral

$$ c_1\int_{-\infty}^{\infty}{\frac{\cos\left(x\tau\right)}{\left(1 + c_{2}\,x^{2}\right)^{\alpha}}}\,
\,{\rm d}x = \frac{ c_{{1}}{c_{{2}}}^{-1/4-1/2\,\alpha}{\pi}^{3/2} \,{\tau}^{\alpha-1/2}{2}^{1/2-\alpha} }{\Gamma \left(\alpha\right) \cos \left( \pi \,\alpha  \right) }
{{\rm I}_{1/2-\alpha}\left({\frac { {\tau}}{\sqrt {c_{{2}}}}}\right)} $$

in terms of the modified Bessel function of the first kind. Pay attention for what values of $\alpha$ the above formula make sense.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
$\ds{\int_{-\infty}^{\infty}{c_{1} \over\pars{1 + c_{2}\,x^{2}}^{5/6}}\,
\cos\pars{x\tau}\,\dd x\,,\qquad c_{1}, c_{2}\mbox{: positive constants.}}$

\begin{align}&\int_{-\infty}^{\infty}{c_{1} \over\pars{1 + c_{2}\,x^{2}}^{5/6}}\,
\cos\pars{x\tau}\,\dd x
=2c_{1}\int_{0}^{\infty}{\cos\pars{x\verts{\tau}} \over\pars{1 + c_{2}\,x^{2}}^{5/6}}\,
\,\dd x
\\[3mm]&={2c_{1} \over c_{2}^{5/6}}\int_{0}^{\infty}
{\cos\pars{x\color{#c00000}{\verts{\tau}}}\over
\bracks{x^{2} + \pars{\color{#c00000}{c_{2}^{-1/2}}}^{2}}^{\color{#c00000}{1/3} + 1/2}}\,\dd x
\end{align}

With the Bessel Function Identity ${\bf 9.6.25}$:
$$
{\rm K}_{\nu}\pars{xz}
={\Gamma\pars{\nu + 1/2}\pars{2z}^{\nu} \over \root{\pi}x^{\nu}}
\int_{0}^{\infty}{\cos\pars{xt} \over \pars{t^{2} + z^{2}}^{\nu + 1/2}}\,\dd t
$$
we'll have
\begin{align}
&\int_{-\infty}^{\infty}{\cos\pars{x\tau} \over\pars{1 + c_{2}\,x^{2}}^{5/6}}\,
\,\dd x
={\root{\pi}\verts{\tau}^{1/3} \over \Gamma\pars{1/3 + 1/2}\pars{2/\root{c_{2}}}^{1/3}}\,
{\rm K}_{1/3}\pars{\verts{\tau}\,{1 \over \root{c_{2}}}}
\end{align}

\begin{align}&\color{#66f}{\large%
\int_{-\infty}^{\infty}{c_{1} \over\pars{1 + c_{2}\,x^{2}}^{5/6}}\,
\cos\pars{x\tau}\,\dd x}
\\[3mm]&=\color{#66f}{\large{c_{1}\root{\pi}\over
\Gamma\pars{5/6}}\,\pars{\verts{\tau}\root{c_{2}} \over 2}^{1/3}
\,{\rm K}_{1/3}\pars{\verts{\tau} \over \root{c_{2}}}}
\end{align}

